When finding the length of the hypotenuse of a right triangle why might you approximate the answer?
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As I read this one, is just a decay exponential equation... so
A = P(1 + r)ᵗ where "t" is days passed. . hmm in this case is decay, so negative rate A = P(1 - r)ᵗ, and the decimal amount would be 0.00877 for the rate
62% of P, the original value, is just 0.62P, now... if we hmm take P as just 1, it could be any amount, but 62% of 1,000,000 is just 62% of 1 times 1,000,000
so, for the sake of comparing it with a percentage, 1 will do
![\bf 0.62P=P(1-0.00877)^t\implies P=1\implies 0.62=(1-0.00877)^t \\\\\\ ln(0.62)=ln[(1-0.00877)^t]\implies ln(0.62)=t\cdot ln(1-0.00877) \\\\\\ \cfrac{ln(0.62)}{ln(0.99123)}=t](https://tex.z-dn.net/?f=%5Cbf%200.62P%3DP%281-0.00877%29%5Et%5Cimplies%20P%3D1%5Cimplies%200.62%3D%281-0.00877%29%5Et%0A%5C%5C%5C%5C%5C%5C%0Aln%280.62%29%3Dln%5B%281-0.00877%29%5Et%5D%5Cimplies%20ln%280.62%29%3Dt%5Ccdot%20%20ln%281-0.00877%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bln%280.62%29%7D%7Bln%280.99123%29%7D%3Dt)
A = P(1 + r)ᵗ where "t" is days passed. . hmm in this case is decay, so negative rate A = P(1 - r)ᵗ, and the decimal amount would be 0.00877 for the rate
62% of P, the original value, is just 0.62P, now... if we hmm take P as just 1, it could be any amount, but 62% of 1,000,000 is just 62% of 1 times 1,000,000
so, for the sake of comparing it with a percentage, 1 will do